1. science
2. mathematics
3. algebra

MAY THE CAUCHY TRANSFORM OF A NON-TRIVIAL FINITE
MEASURE VANISH ON THE SUPPORT OF THE MEASURE ?
XAVIER TOLSA AND JOAN VERDERA
Abstract. Consider a finite complex Radon measure µ in the plane whose
Cauchy transform vanishes µ-almost everywhere on the support of µ. It looks
like, excluding some trivial cases, µ should be the zero measure. We show that
this is the case if certain additional conditions hold.
1. Introduction
Let µ be a finite complex Radon measure in the plane and let C(µ) = z1 ? µ be
its Cauchy Transform. Being the convolution of the locally integrable function z1
with a finite measure, C(µ) is a locally integrable function (with respect to planar
Lebesgue measure dA) and thus it is defined almost everywhere with respect to
dA. Let S be the closed support of the measure µ. For z ∈ S, the value C(µ)(z)
does not need to be defined µ-almost everywhere, because µ may be singular with
respect to dA. We set
Z
dµ(w)
Cε (µ)(z) =
|z−w|>ε z − w
and
Z
dµ(w)
C(µ)(z) = lim Cε (µ)(z) = P.V.
,
ε→0
z−w
whenever the principal value integral exists. Notice that for z ∈
/ S the above limit
exists dA-almost everywhere and coincides with the value of the locally integrable
function C(µ).
Melnikov and Volberg raised the following question. Assume that C(µ) exists
and vanishes µ-almost everywhere on S. Does it then follow that µ = 0?
A first remark is that the answer is obviously no for a point mass. On the other
hand, this is the only example we know that provides a negative answer.
A second remark is that if for any reason C(µ) turns out to be continuous everywhere and S is compact, then the answer is yes, just by the maximum principle.
1
This happens, for instance, if the Newtonian Potential |z|
? µ is continuous and S
is compact; in particular, if µ = f dA with f a compactly supported function in
∈ Lp (C), p > 2.
In this paper we provide two sufficient conditions that ensure that the answer to
our problem is positive. In fact we strongly believe that the answer should be yes
except for the case of atomic measures.
Theorem 1. Let µ = f dA with f a complex valued function in L1 (dA) and assume
that C(f dA) vanishes dA-almost everywhere on the support of f . Then f = 0 dAalmost everywhere.
2000 Mathematics Subject Classification: Primary 30E20.
1
2
XAVIER TOLSA AND JOAN VERDERA
To state the next result we need to introduce some notation and recall some
well-known facts.
Given a positive Radon measure µ in the plane the total Menger curvature (see
[Mel]) of µ is
ZZZ
1
c2 (µ) =
dµ(z)dµ(w)dµ(ζ),
R(z, w, ζ)2
where R(z, w, ζ) is the radius of the disc through z, w and ζ (the inverse of R(z, w, ζ)
is called the Menger curvature of the triple (z, w, ζ)).
The one dimensional fractional maximal function of µ is defined by
M1 µ(z) = sup
r>0
µD(z, r)
,
r
where D(z, r) is the open disc with center z and radius r.
We will also use the standard Hardy-Littlewood maximal operator associated to
area measure, namely
µD(z, r)
M2 µ(z) = sup
πr2
r>0
and the Hardy-Littlewood maximal operator associated to a positive Radon measure µ
Z
1
Mµ f (z) = sup
|f (w)|dµ(w), f ∈ L1loc (µ),
r>0 µ(D(z, r)) D(z,r)
where z is in the support of µ.
µ(D(x, r))
The upper one dimensional density of µ is θµ∗ (x) = lim supr→0
.
r
2
∗
It is a deep theorem of L´eger [L´e] that if c (µ) < ∞ and 0 < θµ (z) < ∞, µ-almost
everywhere, then µ vanishes out of a rectifiable set, that is, out of a countable union
of rectifiable curves.
Using this, it was proved in [To1] that if c2 (µ) < ∞ and M1 µ(z) < ∞ µalmost everywhere, then the principal value integral C(µ)(z) exists µ-almost everywhere. To get readily a more complete and symmetric statement, set θµ (z) =
µ(D(x, r))
limr→0
, whenever the limit exists.
r
We now obtain theR following: if µ is a finite positive Radon measure on C such
that c2 (µ) < ∞ and (M1 µ)2 dµ < ∞, then θµ (z) and the principal value integral
C(µ)(z) exist µ-almost everywhere. Indeed, let F = {z ∈ C : θµ∗ (z) > 0}. From
the results in [L´e] it follows that F is rectifiable. Since M1 µ(x) < ∞, µ-a.e., there
exists a non negative function f such that µ|F is absolutely continuous with respect
to length measure on F and has density f . By the rectifiability of F and Lebesgue’s
differentiation theorem, θµ (z) exists and coincides with f (z) at µ-almost all z ∈ F .
On the other hand, for z 6∈ F we have θµ (z) = 0. The existence of C(µ) in the
principal value sense µ almost everywhere follows now from the results in [MaMe]
and [To1].
The surprising new relevant fact is that we can moreover obtain the following
precise identity.
Theorem
2. Assume that µ is a positive finite Radon measure with c2 (µ) < ∞
R
and (M1 µ)2 dµ < ∞. Then we have
Z
1
π2
2
2
θµ (z)2 dµ(z) + c2 (µ),
(1)
kC(µ)kL2 (µ) = lim kCε µkL2 (µ) =
ε→0
3
6
VANISHING CAUCHY TRANSFORMS
3
where C(µ) is understood in the principal value sense.
In particular, if C(µ) = 0 µ-almost everywhere, then µ = 0.
Notice that the latter statement follows readily from the former, because if C(µ)
vanishes µ-almost everywhere, then the total Menger curvature of µ is zero, which
means that µ is supported on a straight line. Since the density θµ (z) also vanishes
µ-almost everywhere and µ is absolutely continuous with respect to length on that
line, µ must be the zero measure.
Let us remark that the identity (1) has already been proved in [To4, Lemma 6]
in the particular case where µ is the arc length measure on a finite collection of
pairwise disjoint compact segments.
For some recent results concerning principal values of Cauchy integrals, see
[Ma1], [MaP], [To2] and [JP]. See also [Ma2, Section 6] for other related open
problems and [To3] for another connected question.
Section 1 contains the proof of Theorem 1 and section 2 the proof of Theorem
2. We use standard notations and terminology. In particular, the letter C will
denote a constant that does not depend on the relevant parameters involved and
that may vary from one occurrence to another. The notation A . B is equivalent
to A ≤ CB.
2. Proof of Theorem 1
The next lemma is folklore. We include a proof for the reader’s convenience.
Lemma 3. Let ν be a positive finite Radon measure. Then
¶
µ
¯n
o¯
1
kνk2
¯
¯
? ν (w) > t ¯ ≤ C 2 ,
¯ w∈C:
|z|
t
where |E| denotes the area of the set E.
ª
©
¡1
? ν)(w) > t . Then
Proof. Set E = w ∈ C : |z|
µ
¶
¶
Z
Z µ
1
1
1
1
|E| ≤
? ν (w) dA(w) =
? χE dA (ζ) dν(ζ).
|z|
t
|z|
E t
We now take R = |E|1/2 and write:
µ
¶
Z
Z
1
1
1
? χE dA (ζ) =
χE (z)dA(z) +
χE (z)dA(z).
|z|
D(ζ,R) |ζ − z|
C\D(ζ,R) |ζ − z|
The first integral is estimated by 2πR and the second by
|E| ≤ C
|E|
R
= |E|1/2 . Hence
1
|E|1/2 kνk,
t
which proves the Lemma.
Let µ be a complex finite Radon measure and set
Z
1
1
1
dµ(w),
F (z) = C(µ)(z) =
π
π
z−w
µD(z, ε)
µ
e(z) = lim
,
ε→0
πε2
and
Z
1
dµ(w)
Bµ(z) = − P.V.
.
π
(z − w)2
¤
4
XAVIER TOLSA AND JOAN VERDERA
Notice that the above expressions exist dA-almost everywhere. Indeed, µ
e coincides
dA-almost everywhere with the absolutely continuous part of µ (and thus vanishes
if µ is singular with respect to dA) and B(µ) is the Beurling Transform of µ. We
are interested in the differentiability properties of F . Recall that
∂F
∂F
=µ
and
= Bµ,
∂z
∂z
in the sense of distributions. Thus we set for w 6= z
Q(w, z) = Qµ (w, z) =
|F (w) − F (z) − Bµ(z)(w − z) − µ
e(z)(w − z)|
.
|w − z|
It is not true that F is differentiable at dA-almost all points z, but the following
weaker substitute result is available (the method is inspired by [St, Chapter 8]).
Lemma 4. For dA-almost all z ∈ C we have
1
lim sup t |{w ∈ D(z, ε) : Q(w, z) > t}|1/2 = 0.
ε→0 ε t>0
Proof. Notice that the conclusion of the Lemma holds if µ is of the form ϕ dA, with
ϕ a compactly supported infinitely differentiable function, or if µ is supported on a
closed set of zero area. Therefore the set of measures for which the conclusion of the
Lemma holds is dense in the space of all finite complex Radon measures endowed
with the total variation norm. Consider the sub-linear operator
1
T µ(z) = sup sup t |{w ∈ D(z, ε) : Qµ (w, z) > t}|1/2 .
ε>0 ε t>0
By the opening remark it is clearly enough to show that T satisfies the weak type
inequality
C
|{z ∈ C : T µ(z) > t}| ≤ kµk,
t
which follows from
(2)
T µ(z) ≤ C {M2 µ(z) + B ∗ µ(z)}, z ∈ C,
where B ∗ stands for the maximal Beurling transform, that is,
B ∗ µ(z) = sup |Bε µ(z)|
ε>0
and
1
Bε µ(z) =
π
Z
|w−z|>ε
dµ(w)
.
(z − w)2
To prove (2) take z = 0, w 6= 0 and set δ = |w|. Then
|F (w) − F (0) − Bµ(0)w − µ
e(0)w|
≤ |F (w) − F (0) − B2δ µ(0)w| + |w||B2δ µ(0) − Bµ(0)| + |w|M2 µ(0)
≤ |F (w) − F (0) − B2δ µ(0)w| + 2|w|B ∗ µ(0) + |w|M2 µ(0)
and, since we can assume without loss of generality that µ is a positive measure,
¯
¯
Z
¯ 1
1
1 ¯¯
¯
|F (w) − F (0) − B2δ µ(0)w| ≤
+ ¯ dµ(ζ)
π |ζ|<2δ ¯ w − ζ
ζ
¯
¯
Z
¯
¯
1
¯ 1 + 1 + w ¯ dµ(ζ) = I + II.
+
¯
2
π |ζ|>2δ w − ζ
ζ
ζ ¯
VANISHING CAUCHY TRANSFORMS
5
We first estimate the term II. Observe that if |ζ| > 2δ then
¯
¯
¯ 1
1
w ¯¯
|w|2
|w|2
¯
+
+
≤
2
.
≤
¯w − ζ
ζ
ζ 2 ¯ |w − ζ||ζ|2
|ζ|3
Thus
Z
π II ≤
2
|ζ|>2δ
|w|2
dµ(ζ) ≤ C|w|M2 µ(0),
|ζ|3
where in the last inequality we use the standard idea of decomposing the domain
of integration in annuli centered at 0 whose distance to 0 is of the order of δ 2n .
The term I is π times the integral
µ
¶
Z
|w|
1
dµ
dµ(ζ) = |w|
? χD(0,2δ)
(w).
|ζ|
|ζ|
|ζ|<2δ |w − ζ||ζ|
Since δ = |w| we obtain using Lemma 3
¯½
µ
¶
¾¯1/2
¯
1
t ¯¯
dµ
¯
w
∈
D(0,
ε)
:
?
χ
(w)
>
t
D(0,2δ)
¯
¯
ε
|ζ|
|ζ|
¯½
µ
¶
¾¯1/2
¯
t ¯¯
1
dµ
≤ ¯ w ∈ D(0, ε) :
? χD(0,2ε)
(w) > t ¯¯
ε
|ζ|
|ζ|
Z
dµ(ζ)
C
≤ CM2 µ(0),
≤
ε D(0, 2ε) |ζ|
where in the last inequality we decompose again the domain of integration in annuli
centered at 0 whose distance to 0 is of the order of ε 2−n .
¤
Proof of Theorem 1. Let S be the closed support of the function f and set E =
{z ∈ S : F (z) = 0 and f (z) 6= 0}, so that we have to show that |E| = 0. We
will prove that almost all points of E are not points of density of S. Take a point
z ∈ E which is a Lebesgue point of f and at which the conclusion of Lemma 4
holds. Assume that z = 0 and consider the R-linear mapping L : C → C defined
by L(w) = aw + bw, where a = Bµ(0), µ = f dA and b = f (0). Since b 6= 0 L is not
identically 0. We distinguish two cases.
Case 1 : the kernel of L is {0}. Then for some δ > 0 we have
(3)
|aw + bw|
≥ δ, |w| = 1.
|w|
Write, as in the proof of Lemma 4, F (z) =
Q(w) =
1
π
C(µ)(z), where µ = f dA, and
|F (w) − aw − bw|
, w 6= 0.
|w|
Take t = δ/2 (in fact, any other any positive number t less than δ would work) .
Taking into account that F vanishes on S dA-almost everywhere and (3), we obtain
t2 |{w ∈ D(0, ε) ∩ S : Q(w) > t}|
t2 |{w ∈ D(0, ε) : Q(w) > t}|
=
2
ε
ε2
2
t |{w ∈ D(0, ε) \ S : Q(w) > t}|
+
ε2
2
t |D(0, ε) ∩ S|
≥
.
ε2
6
XAVIER TOLSA AND JOAN VERDERA
Thus Lemma 4 tells us that |D(0,ε)∩S|
tends to 0 with ε, which means that 0 is not
ε2
a point of density of S.
Case 2 : the kernel of L is one dimensional. Assume without loss of generality
that L(1) = 0. Then
√ L(w) = aw − aw = a2i Im(w). Thus, on the cone K = {w ∈
C : |Im(w)| ≥ |w|/ 2}, we have the inequality
|aw + bw|
|Im(w)| √
= 2|a|
≥ 2|a| = δ > 0,
|w|
|w|
where the last identity is a definition of δ. Take now any positive number t less
than δ. As before we have
t2 |{w ∈ D(0, ε) : Q(w) > t|
t2 |D(0, ε) ∩ S ∩ K|
≥
ε2
ε2
and therefore by Lemma 4
lim
ε→0
|D(0, ε) ∩ S ∩ K|
= 0.
ε2
On the other hand
|D(0, ε) ∩ S \ K|
|D(0, ε) \ K|
1
≤
≤ ,
2
2
πε
πε
2
and hence 0 is not a point of density of S.
¤
3. Proof of Theorem 2
To prove Theorem 2 we will need some preliminary lemmas. The first one is the
following well known estimate.
Lemma 5. Let µ be a positive Radon measure on C, and z ∈ C. For 0 < r1 <
r2 ≤ ∞, we have
Z
1
C
µ(D(z, R))
M1 µ(z)
dµ(w) ≤
sup
≤C
.
2
r1 r1 ≤R≤r2
R
r1
r1 ≤|z−w|≤r2 |z − w|
Next lemma is also easy.
Lemma 6. Let µ be a positive Radon measure on C, and z ∈ C such that θµ (z) = 0.
Then,
Z
1
lim r
dµ(w) = 0.
r→0
|z
−
w|2
|z−w|≥r
Proof. Let K > 1 be some big constant. For any r > 0, by the preceding lemma
we have
Z
Z
Z
r
r
r
dµ(w)
=
dµ(w)
+
dµ(w)
2
2
2
|z−w|≥r |z − w|
r≤|z−w|≤Kr |z − w|
|z−w|>Kr |z − w|
.
sup
r≤R≤Kr
As a consequence,
Z
lim sup
r→0
|z−w|≥r
µ(D(z, R))
1
+
M1 µ(z).
R
K
1
r
dµ(w) .
M1 µ(z).
|z − w|2
K
Since M1 µ(z) < ∞ and K is arbitrary, the lemma follows.
¤
VANISHING CAUCHY TRANSFORMS
7
The following version of Cotlar’s inequality has been proved in [Vo, Lemma 5.1]
(to be precise, there are some little differences between [Vo, Lemma 5.1] and the
following result; however the arguments are almost the same):
Lemma 7. Let µ be a positive Radon measure on C. Then, at µ-a.e. z ∈ C we
have
sup |Cε µ(z)| ≤ C {Mµ (Cδ µ)(z) + M1 µ(z)},
ε≥δ
where the constant C does not depend on δ.
Notice that in this lemma one needs not to assume the L2 (µ) boundedness of
the Cauchy integral operator.
Recall the identity proved in [MeV]:
ZZZ
1
2
(4)
kCε µk2L2 (µ) =
|z−w|>ε c(z, w, ζ) dµ(z)dµ(w)dµ(ζ)
6
|z−ζ|>ε
|w−ζ|>ε
ZZZ
+
1
|z−w|≤ε
|z−ζ|>ε
|w−ζ|>ε
(ζ − z)(ζ − w)
dµ(z)dµ(w)dµ(ζ).
R
It is well known that the last integral in (4) is bounded above by C (M1 µ)2 dµ.
Indeed, for z, w, ζ such that |z − w| ≤ ε, |z − ζ| > ε and |w − ζ| > ε we have
|z − ζ| ≈ |w − ζ|, and so by Lemma 5, for any given z we get
ZZ
ZZ
1
1
dµ(w)dµ(ζ) . |z−w|≤ε
dµ(w)dµ(ζ)
|z−w|≤ε
2
|z−ζ|>ε |ζ − z| |ζ − w|
|z−ζ|>ε |ζ − z|
|w−ζ|>ε
|w−ζ|>ε
Z
≤ µ(D(z, ε))
|z−ζ|>ε
1
dµ(ζ)
|ζ − z|2
M1 µ(z)
. µ(D(z, ε))
(5)
≤ M1 µ(z)2 ,
ε
and our claim follows. So by the assumptions in the theorem, it turns out that
kCε µkL2 (µ) is bounded uniformly on ε > 0. As a consequence, by the preceding
version of Cotlar’s inequality and monotone convergence, we have kC ∗ µkL2 (µ) < ∞,
where C ∗ µ(z) = supε>0 |Cε µ(z)|. By dominated convergence this implies
kCµkL2 (µ) = lim kCε µkL2 (µ) < ∞.
ε→0
The identity (4) also tells us that
ZZZ
1
1
lim kCε µk2L2 (µ) = c2 (µ) + lim
dµ(z)dµ(w)dµ(ζ).
|z−w|≤ε
ε→0
ε→0
6
(ζ
−
z)(ζ
− w)
|z−ζ|>ε
|w−ζ|>ε
Therefore, in order to prove the theorem we only have to show that the triple
2 R
integral
on the right side above converges to π3 θµ (z)2 dµ. By the assumption
R
M1 µ2 dµ < ∞, (5), and dominated convergence, to prove this statement it suffices
to show that at µ-a.e. z ∈ C we have
ZZ
1
π2
(6)
lim |z−w|≤ε
dµ(w)dµ(ζ) =
θµ (z)2 .
ε→0
3
(ζ
−
z)(ζ
−
w)
|z−ζ|>ε
|w−ζ|>ε
8
XAVIER TOLSA AND JOAN VERDERA
If θµ (z) = 0 this is easy: from (5) we get
ZZ
1
M1 µ(z)
dµ(w)dµ(ζ) . µ(D(z, ε))
,
|z−w|≤ε
|ζ
−
z|
|ζ
−
w|
ε
|z−ζ|>ε
|w−ζ|>ε
which tends to 0 because M1 µ(z) < ∞.
Suppose now that θµ (z) > 0. In this case z belongs to the rectifiable set F
introduced above. We want to reduce the problem to the case where µ is supported
on a Lipschitz graph. To this end, recall that there is aScountable family of Lipschitz
graphs Γn and a set Z with µ(Z) = 0 such that F ⊂ n Γn ∪ Z. Let z be a density
point of Γn (i.e. µ(D(z, r) \ Γn )/µ(D(z, r)) → 0 as r → 0). Let us see that
ZZ
1
(7)
lim |z−w|≤ε
dµ(w)dµ(ζ) = 0.
ε→0
(ζ − z)(ζ − w)
|z−ζ|>ε
|w−ζ|>ε
w6∈Γn or ζ6∈Γn
By estimates analogous to the ones in (5) it is straightforward to check that
ZZ
1
dµ(w)dµ(ζ) = 0.
(8)
lim |z−w|≤ε
ε→0
1/2 |ζ − z||ζ − w|
|z−ζ|>ε
|w−ζ|>ε
So we may assume that |z − ζ| ≤ ε1/2 in the integral in (7). We consider first the
case w 6∈ Γn :
ZZ
Z
1
1
dµ(w)dµ(ζ)
.
µ(D(z,
ε)
\
Γ
)
dµ(ζ)
n
|z−w|≤ε
|z
−
ζ|2
1/2 |ζ − z||ζ − w|
|z−ζ|>ε
ε<|z−ζ|≤ε
|w−ζ|>ε
w6∈Γn
≤ µ(D(z, ε) \ Γn )
≤
M1 µ(z)
µ(D(z, ε) \ Γn ) µ(D(z, ε))
=
M1 µ(z)
ε
µ(D(z, ε))
ε
µ(D(z, ε) \ Γn )
M1 µ(z)2 → 0 as ε → 0,
µ(D(z, ε))
assuming M1 µ(z) < ∞. The case ζ 6∈ Γn is similar. The details are left to the
reader.
By (7), when proving (6) for z ∈ F , we may assume that µ is supported on a
Lipschitz graph Γn . We show that (6) holds in this case in a separate lemma.
Lemma 8. Let Γ be a Lipschitz graph and µ a Radon measure supported on it such
1
that µ = g dH|Γ
and µ(Γ) < ∞. Then,
ZZ
(9)
lim
ε→0
1
|z−w|≤ε
|z−ζ|>ε
|w−ζ|>ε
(ζ − z)(ζ − w)
dµ(w)dµ(ζ) =
π2
g(z)2
3
at µ-a.e. z ∈ Γ.
1
.
Proof. First we consider the case where Γ coincides with the real line and µ = H|R
The proof of (9) in this situation is basically contained in [To4, Lemma 7]. However,
for completeness we will also show the detailed calculations here. We may assume
VANISHING CAUCHY TRANSFORMS
9
that z = 0. By (8), showing that (9) holds is equivalent to proving that
ZZ
1
π2
1
Iε :=
dy dz →
as ε → 0,
|y|≤ε
z(z − y)
3
ε<|z|≤ε1/2
|y−z|>ε
where dy and dz denote the usual integration with respect to Lebesgue measure in
R. On the other hand, by symmetry it is easy to check that
ZZ
1
1
Iε = 2 0<y≤ε
dy dz.
1/2 z(z − y)
ε<|z|≤ε
|y−z|>ε
Thus
Z ε µZ
¶
1
=2
dz dy.
0
−ε1/2
y+ε z(z − y)
¯
¯
¯
¯
Since a primitive of 1/(z(z − y)) (with respect to z) is y1 log¯1 − yz ¯, it follows that
¸
Z ε·
¯
¯
¯
2
y ¯¯ 1
y ¯¯ 1
y ¯¯
¯
¯
¯
1
Iε = 2
log¯1 + ¯ + log¯1 − 1/2 ¯ − log¯1 + 1/2 ¯ dy.
y
ε
y
y
ε
ε
0
Iε1
−ε
1
dz +
z(z − y)
Z
ε1/2
If we split the integral into two parts and we change variables, we get
Z 1
Z ε1/2
¢
1
1¡
1
(10)
Iε = 4
log |1 + t| dt + 2
log |1 − t| − log |1 + t| dt.
t
t
0
0
It is well known that
Z
1
π2
1
log |1 + t| dt =
.
12
0 t
On the other hand, the last integral in (10) tends to 0 as ε → 0 because the function
inside the integral is bounded in (0, 1/2]. Thus Iε1 → π 2 /3 as ε → 0, and so the
1
lemma holds for Γ = R and µ = H|R
.
Consider now the case of a general Lipschitz graph. Let A be a Lipschitz function
such that Γ = {(s, A(s)) : s ∈ R}. Assume that z ∈ Γ is a Lebesgue point
1
of g with respect to H|Γ
, that M1 µ(z) < ∞, that (s, A(s)) = z and that A is
differentiable at s. Moreover, without loss of generality, we may suppose also that
0
s=
A0 , since the term
£ 0 and A(0) =¤A (0) = 0 and that 0 is a Lebesgue point of
1
1/ (ζ − z)(ζ − w) is invariant by translations and rotations of z, w, ζ. By (8),
showing that (9) holds for this z is equivalent to proving that
ZZ
1
π2
lim |w|≤ε
dµ(w)dµ(ζ) =
g(0)2 .
ε→0
3
1/2 ζ(ζ − w)
ε<|ζ|≤ε
|w−ζ|>ε
To prove this we will compare the integral above with
ZZ
1
2
g(0)
ds dt,
|s|≤ε
t(t
−
s)
ε<|t|≤ε1/2
|s−t|>ε
1One needs a rotation in order to get A0 (0) = 0. However a little problem (easy to fix) may
appear: when one rotates a Lipschitz graph one obtains the Lipschitz graph of another function
if the rotation angle is small enough. To ensure that the rotation angle is small, if necessary one
can assume that the Lipschitz constant of the graph is small, or even one can suppose that A0 is
continuous and then argue locally, etc.
10
XAVIER TOLSA AND JOAN VERDERA
where ds and dt denote the usual Lebesgue measure on R. The latter integral
converges to π 2 /3, as shown above. Let us denote γ(s) = s + i A(s). We have
¯Z Z
¯
¯ |w|≤ε
¯
ZZ
1
ε<|ζ|≤ε1/2
|w−ζ|>ε
¯Z Z
¯
≤ ¯¯ |w|≤ε
ζ(ζ − w)
dµ(w)dµ(ζ) − g(0)2
ZZ
1
ε<|ζ|≤ε1/2
|w−ζ|>ε
ζ(ζ − w)
¯Z Z
¯
+ ¯¯ |w|≤ε
dµ(w)dµ(ζ) −
ζ(ζ − w)
¯Z Z
¯
+ ¯¯ |w|≤ε
|w|≤ε
ε<|ζ|≤ε1/2
|w−ζ|>ε
ζ(ζ − w)
1
dH|Γ
(w)dµ(ζ) −
¯
¯
1
(w)dµ(ζ)¯¯
dH|Γ
g(0)2
|w|≤ε
ε<|ζ|≤ε1/2
|w−ζ|>ε
ζ(ζ − w)
ZZ
g(0)2
ε<|ζ|≤ε1/2 ζ(ζ
|w−ζ|>ε
g(0)
ZZ
g(0)
ε<|ζ|≤ε1/2
|w−ζ|>ε
|s|≤ε
ε<|t|≤ε1/2
|s−t|>ε
¯
¯
1
ds dt¯¯
t(t − s)
− w)
1
1
dH|Γ
(w)dH|Γ
(ζ) −
¯
¯
1
1
dH|Γ
(w)dH|Γ
(ζ)¯¯
g(0)2
|γ(s)|≤ε
ε<|γ(t)|≤ε1/2 γ(t)(γ(t)
|γ(s)−γ(t)|>ε
− γ(s))
¯
¯
ds dt¯¯
¯
¯
¯
1
1 ¯¯
¯
ds dt
−
|γ(s)|≤ε
¯
t(t − s) ¯
ε<|γ(t)|≤ε1/2 γ(t)(γ(t) − γ(s))
ZZ
+ g(0)2
|γ(s)−γ(t)|>ε
¯Z Z
¯
2¯
+ g(0) ¯ |γ(s)|≤ε
ε<|γ(t)|≤ε1/2
|γ(s)−γ(t)|>ε
1
ds dt −
t(t − s)
ZZ
|s|≤ε
ε<|t|≤ε1/2
|s−t|>ε
¯
¯
1
ds dt¯¯
t(t − s)
=: D1 + . . . + D5 .
We will show that each term D1 , . . . , D5 tends to 0 as ε → 0, and we will be done.
Let us deal with D1 first:
ZZ
1
1
|g(w) − g(0)| dH|Γ
(w)dµ(ζ)
D1 .
|w|≤ε
2
1/2 |ζ|
ε<|ζ|≤ε
|w−ζ|>ε
µ Z
≤ ε
|ζ|>ε
¶µ Z
¶
1
1
1
dµ(ζ)
|g(w) − g(0)| dH|Γ (w) .
|ζ|2
ε |w|<ε
The first factor on the right side is bounded above by M1 µ(0), and the last factor
tends to 0 because 0 is a Lebesgue point of g. Thus D1 → 0 as ε → 0.
We now consider the term D2 . Using again the fact that 0 is a Lebesgue point
of g, we get
ZZ
D2 .
|w|≤ε
ε<|ζ|≤ε1/2
|w−ζ|>ε
g(0)
1
1
|g(ζ) − g(0)| dH|Γ
(w)dH|Γ
(ζ)
|ζ|2
Z
. εg(0)
ε<|ζ|≤ε1/2
. εg(0)
1
ε
sup
0<r<ε1/2
|g(ζ) − g(0)|
1
dH|Γ
(ζ)
|ζ|2
Z
1
1
|g(ζ) − g(0)| dH|Γ
(ζ) → 0
r |ζ|≤r
as ε → 0,
1
where we applied Lemma 5 to measure |g(ζ) − g(0)| dH|Γ
(ζ) in the last inequality.
VANISHING CAUCHY TRANSFORMS
11
Let us turn our attention to D3 . We have
ZZ
¯
1 ¯¯
(11) D3 . g(0)2 |γ(s)|≤ε
(1 + A0 (s)2 )1/2 (1 + A0 (t)2 )1/2 − 1¯ ds dt.
2
γ(t)
ε<|γ(t)|≤ε1/2
|γ(s)−γ(t)|>ε
Since A0 is bounded, we get
¯
¯
¯(1 + A0 (s)2 )1/2 (1 + A0 (t)2 )1/2 − 1¯
¯
¯
¯
¯
≤ ¯(1 + A0 (s)2 )1/2 − 1¯ (1 + A0 (t)2 )1/2 + ¯(1 + A0 (t)2 )1/2 − 1¯
. |A0 (s)| + |A0 (t)|.
We wish to replace the conditions on w = γ(s) and ζ = γ(t) in the domain of
integration of D3 by similar conditions on s and t. To this end, notice that w =
s (1 + o(1)) and ζ = t (1 + o(1)) as ε → 0 since A0 (0) = 0 and |s| ≤ ε, |t| ≤ ε1/2 . As
a consequence,
¯
¯
¯|w − ζ| − |s − t|¯ ≤ |w − s| + |ζ − t| ≤ (|w| + |ζ|)o(1) ≤ 2|ζ|o(1) . |w − ζ|o(1)
and so |w − ζ| = |s − t| (1 + o(1)). Therefore, for ε small enough we have
ZZ
1
2
D3 . g(0)
(|A0 (s)| + |A0 (t)|) ds dt
|s|≤2ε
2
t
ε/2<|t|≤2ε1/2
g(0)
.
ε
2
g(0)
ε
2
.
Z
|s−t|>ε/2
Z
0
|A (s)| ds + εg(0)
2
ε/2<|t|≤2ε1/2
|s|≤2ε
Z
|A0 (s)| ds + g(0)2
sup
0<r≤2ε1/2
|s|≤2ε
1
r
Z
|A0 (t)|
dt
t2
|A0 (t)| dt.
|t|≤r
Since A0 (0) = 0 and 0 is a Lebesgue point of A0 , we deduce that D3 → 0 as ε → 0.
Let us consider D4 . For s, t and w = γ(s), ζ = γ(t) as in the integral in D4 we
have
¯ ¯
¯ ¯
¯
¯
¯ ¯
¯
1 ¯¯ ¯¯
1
1 ¯¯
1
1
1
¯+¯
¯
−
≤
−
−
¯ ζ(ζ − w) t(t − s) ¯ ¯ ζ(ζ − w) t(ζ − w) ¯ ¯ t(ζ − w) t(t − s) ¯
|ζ − t| + |w − s|
o(1)
|ζ − t|
+
.
.
≤
|ζ − w| |ζ| |t| |t| |ζ − w| |t − s|
|t|2
Thus,
ZZ
D4 . g(0)2
|s|≤2ε
ε/2<|t|≤2ε1/2
|s−t|>ε/2
o(1)
ds dt . g(0)2 o(1).
t2
So D4 also tends to 0 when ε → 0.
Finally we estimate D5 . Let S be the symmetric difference between
{(s, t) ∈ R2 : |w| < ε, ε < |ζ| ≤ ε1/2 , |w − ζ| > ε}
and
{(s, t) ∈ R2 : |s| < ε, ε < |t| ≤ ε1/2 , |s − t| > ε}.
12
XAVIER TOLSA AND JOAN VERDERA
RR
We have D5 . g(0)2 S t12 ds dt. We split the integral as follows:
ZZ
ZZ
ZZ
1
···
ds dt ≤
|s|≤2ε
{|w|≤ε}4{|s|≤ε} · · · +
2
S t
{|ζ|>ε}4{|t|>ε}
ε/2<|t|≤2ε1/2
|s−t|>ε/2
|s−t|>ε/2
ZZ
+
ZZ
···
|s|≤2ε
|t|>ε/2
{|ζ|≤ε1/2 }4{|t|≤ε1/2 }
|s−t|>ε/2
+
···
|s|≤2ε
|t|>ε/2
{|s−t|>ε}4{|w−ζ|>ε}
= I1 + · · · + I4 .
Let us deal with I1 :
³
´Z
I1 . H1 {s : |γ(s)| ≤ ε}4{s : |s| ≤ ε}
|t|>ε/2
1
dt
t2
³
´
1
≈ H1 {s : |γ(s)| ≤ ε}4{s : |s| ≤ ε} .
ε
Since |s − γ(s)| = |s|o(1) ≤ ε o(1), we have
³
´
H1 {s : |γ(s)| ≤ ε}4{s : |s| ≤ ε} ≤ ε o(1),
and I1 → 0 as ε → 0. Consider now I2 :
Z
I2 . ε
{|ζ|>ε}4{|t|>ε}
1
dt.
t2
Since |ζ| = |γ(t)| = |t|(1 + o(1)), we have
¡
¢
¡
¢
{t : |γ(t)| > ε}4{t : |t| > ε} ⊂ B 0, ε(1 + o(1)) \ B 0, ε(1 − o(1)) =: A(ε).
In the annulus A(ε) we have |t| ≈ ε. Moreover H1 (R ∩ A(ε)) ≈ ε o(1). So we infer
that I2 . o(1).
The estimates for the integrals I3 and I4 are analogous.
¤
ACKNOWLEDGEMENT. Both authors were supported by the grants MTM
2004-00519, Acci´on Integrada HF2004-0208 (Ministerio de Ciencia y Tecnolog´ıa),
and SGR 2005 (Generalitat de Catalunya).
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´ Catalana de Recerca i Estudis Avanc
Institucio
¸ ats (ICREA) and Departament de
` tiques, Universitat Auto
` noma de Barcelona, Spain
Matema
E-mail address: [email protected]
` tiques, Universitat Auto
` noma de Barcelona, Spain
Departament de Matema
E-mail address: [email protected]